# Density function question

Suppose $x_{1}, x_{2} \dots x_{N}$ are gaussian RVs with variance $S$ and mean $1$. What is the density function of

$$\frac{ |\sum_{n=1}^{N}x_{n}|^{2}}{\sum_{n=1}^{N}|x_{n}|^{2}}\text{?}$$

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Can you tel us why you need the density ? –  robin girard Jun 1 '11 at 19:36
It does not depend upon $S$. –  NRH Jun 1 '11 at 19:46
@NRH actually it depends upon the ratio $\mu/S$, so the density is independent of $S$ only if the mean is zero. –  Thies Heidecke Jun 4 '11 at 12:02
@Theis, thanks for correcting me. You are completely right! –  NRH Jun 4 '11 at 17:33

For the case $N=3,\;x_i\sim\mathcal{N}(\mu,\sigma),\;i\in\{1,2,3\}$, i calculated a CDF of
$P(y\leq\alpha)=\text{exp}\Big(-\frac{1}{2}(3-\alpha)\big(\frac{\mu}{\sigma}\big)^2\Big)\sqrt{\alpha/3}\;,\quad y=\frac{(x_1+x_2+x_3)^2}{x_1^2+x_2^2+x_3^2},\quad 0\leq\alpha\leq N$.
The other cases $N\neq 3$ are much harder to solve (at least for me). An idea that helped the solution was rotating the coordinate system so that the former $(x_1, x_2, x_3)\propto(1,1,1)$-direction is aligned to an axis from the new coordinate system, e.g. $(z_1, z_2, z_3)\propto(0,0,1)$. Then
$y=\frac{(x_1+x_2+x_3)^2}{x_1^2+x_2^2+x_3^2}=\frac{3 z_3^2}{z_1^2+z_2^2+z_3^2}$,